3. Drilling Fluids

Table of Content

  1. Increase Mud Weight
  2. Dilution
  3. Mixing Fluids of Different Densities
  4. Oil Based Mud Calculations
  5. Solids Analysis
  6. Solids Fractions
  7. Dilution of Mud System
  8. Displacement - Barrels of Water/Slurry Required
  9. Evaluation of Hydrocyclone
  10. Evaluation of Centrifuge

3.3 Mixing Fluids of Different Densities

(V1 × D1) + (V2 × D2) = VF × DF

where
V1 = volume of fluid 1 (bbl, gal, etc.)
D1 = density of fluid 1 (ppg,lb/ft3, etc.)
V2 = volume of fluid 2 (bbl, gal, etc.)
D2 = density of fluid 2 (ppg,lb/ft3, etc.)
VF = volume of final fluid mix
DF = density of final fluid mix

Example 1:
A limit is placed on the desired volume:
Determine the volume of 11.0 ppg mud and 14.0 ppg mud required to build 300 bbl of 11.5 ppg mud:
Given: 400 bbl of 11.0 ppg mud on hand, and 400 bbl of 14.0 ppg mud on hand

Solution:
let V1 = bbl of 11.0 ppg mud
V2 = bbl of 14.0 ppg mud
then
a. V1 + V2 = 300 bbl
b. (V1 × 11.0) + (V2 × 14.0) = 300 × 11.5
Multiply Equation (a) by the density of the lowest mud weight (D1 = 11.0 ppg) and subtract the result from Equation (b):
   (b) (V1 × 11.0) + (V2 × 14.0) = 3450
− (a) (V1 × 11.0) + (V2 × 11.0) = 3300
-----------------------------------------------
                        3V2 = 150
therefore,
V2 = 50 bbl of 14.0 ppg mud
V1 = 300 - 50 = 250 bbl of 11.0 ppg mud
Check:
(250 bbl × 11.0 ppg) + (50 bbl × 14.0 ppg) = 300 bbl × 11.5 ppg
2750 + 700 = 3450

Example 2:
No limit is placed on volume:
Determine the density and volume when the two following muds are mixed together:
Given: 400 bbl of 11.0 ppg mud, and 400 bbl of 14.0 ppg mud

Solution:
let V1 = 400 bbl of 11.0 ppg mud
V2 = 400 bbl of 14.0 ppg mud
then
VF = V1 + V2 = 400 bbl + 400 bbl = 800 bbl
and
(V1 × D1) + (V2 × D2) = VF × DF
(400 bbl × 11.0 ppg) + (400 bbl × 14.0 ppg) = 800 bbl × DF
4400 + 5600 = 800 bbl × DF
10,000 = 800 bbl × DF
therefore,
DF = 12.5 ppg
Check:
(400 bbl × 11.0 ppg) + (400 bbl × 14.0 ppg) = 800 bbl × 12.5 ppg
4400 + 5600 = 10,000

⬆ Table of Content